a mixture of KBr ,NaBr weighing 0.56 gm was treated with aqueous solution of Ag+ and the bromide ion was recovered as 0.97 gm of pure AgBr. What was the weight of KBr in the sample
Answers
Answer:
Weight of KBr is equal to 0.2298 g.
Explanation:
Let mass of KBr be be x
Let the mass of NaBr be (0.56 g -x)
Moles of KBr
Moles of NaBr =
Moles of AgBr =
1 mol of KBr gives 1 mole of bromide ions.
1 mol of NaBr gives 1 mole of bromide ions.
Moles of bromides in the solution,n =
If all the bromide ions present in the mixture are recovered as AgBr, then moles of bromide ions will be equal to the moles of AgBr.
x = 0.2298 g
Weight of KBr is equal to 0.2298 g.
weight of NaBr is equal to 0.3302 g
Answer:
Explanation:
Let mass of KBr be be x
Let the mass of NaBr be (0.56 g -x)
Moles of KBr
Moles of NaBr =
Moles of AgBr =
1 mol of KBr gives 1 mole of bromide ions.
1 mol of NaBr gives 1 mole of bromide ions.
Moles of bromides in the solution,n =
If all the bromide ions present in the mixture are recovered as AgBr, then moles of bromide ions will be equal to the moles of AgBr.
x = 0.2298 g
Weight of KBr is equal to 0.2298 g.
weight of NaBr is equal to 0.3302 g
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